Joker Games has released the extremely nice Paso
Doble. You control a puppet that must always take two steps. Very interesting.

In response to the Don Woods puzzle, Bob Kraus has created an
Eight Question Test: Answer each of the 8 questions with a letter from
A to D. The word "answer" in the test refers to YOUR answer, not
some "best" answer. Your score is the number of correct answers.
Try to get the highest possible score. Answers.

(1) The next question with the same answer as this one is:

(A) 2 (B) 3 (C) 4 (D) 5

(2) The first question with answer C is:

(A) 1 (B) 2 (C) 3 (D) 4

(3) The last question with answer A is:

(A) 5 (B) 6 (C) 7 (D) 8

(4) The number of questions with answer D is:

(A) 1 (B) 2 (C) 3 (D) 4

(5) The answer occuring the most is: (if tied, first alphabetically)

(A) A (B) B (C) C (D) D

(6) The first question with the same answer as the question following
it is:

I recently picked up Sort
1.2 for my PDA. You can see it with the Palm
OS Emulator. A java version of much the same thing is at the Sorting
Algorithms page. I perused Art of Computer Programming: Sorting and Searching,
and was very surprised by the list of best known sorting techniques. The following
is a list of commands that show how 9 items can be sorted with 25 comparison-swaps:

I used the "zero-one principle" to prove the nine25
list was valid: If a network with n input lines sorts all 2^n sequences
of 0s and 1s into nondecreasing order, it will sort any arbitrary sequence
of n numbers into non-decreasing order. I used Mathematica to construct
all 512 sequences, and then used the Sorter routine to sort them all. This
network for 9 terms was found by R. W. Floyd in 1964, and has not been
proven optimal. The last effort to prove optimal sort sequences stopped
at 8 terms in 1966. No results above 8 have been proven! Here is the Best
Known list. The Batcher
Method is the best currently known for 17 and up.

I was curious about the importance of each swap in the Floyd
list, so I set qq to the sorted lists, and did this:Table[Length[Flatten[Map[Position[Map[Sorter[#,
Drop[nine25, {n}]] &, onesandzeros[9]], #] &,
Complement[Union[Map[Sorter[#, Drop[nine25, {n}]] &, onesandzeros[9]]],
qq]]]], {n, 1, 25}]

So only 7 sequences don't get sorted correctly when the 12th
term {1,4} is dropped. Hughes Juille, in 1995, found a 45-swap method for
13 items, using a computer program simulating an evolutionary process of genetic
breeding. If you can improve or prove any of the sorting techniques, write
me.

material added 26 November 2001

Although it has a Petersen minor, the Coxeter graph is still
3-colorable. The trick is to find two 14-cycles, which isn't easy. I toyed
around with various presentations. Jim Boyce and Joseph DeVincentis sent solutions.

I've moved images and commentary to my new Sam
Loyd page. A portfolio of John Whiting's art mazes is at knottedlines.com.
Patrick Hamlyn demonstrates that the hexacubes and a 2x2 block can make a
size 10 cube in a lot of ways (available at Kadon).

material added 19 November 2001

Many people know of the Four
Color Theorem. Less known is the Edge Coloring Theorem. Tait, 1880: The
edges of a bridgeless cubic planar graph are three-colorable. A bridged
graph () is not three-colorable.
A cubic graph is one where each vertex has three edges. Three-colorable means
that no two edges of the same color touch. The proof is easy. First, four-color
the map. This can be done via the Four Color Theorem. Next, color cyan any
edge between red and blue regions, or between yellow and green regions. Color
orange any edge between red and green regions, or between blue and yellow
regions. Color magenta any edge between red and yellow regions, or between
blue and green regions. That's it -- you now have a 3-colored edge
coloring.

If you have an edge-colored map, finding the 4 coloring is just
as easy. Remove one of the three colors -- magenta, for example. You are left
with various closed cycles that connect every vertex. Each region will have
an even or odd number of fences around it. Shade the odd-fenced regions red.
Next, do the same thing with a different edge color, like orange. Shade the
odd-fenced regions yellow. Overlay the two pictures, and you get a four-coloring
in red, yellow, orange, and white. Of course, it might not be easy to find
the initial 4-coloring. In 1976, Haken and Appel needed a computer to 4-color
the following hardest-case map, which I present in a slightly different form.

The 3-edge-coloring is known as the Tait-coloring. In 1884,
Tait conjectured that every polyhedra (or 3-connected planar graph) has a
Hamiltonian
cycle. If true, the 4 color theorem follows immediately. A given map can
be made cubic by drawing small circles around each vertex with degree greater
than 3, and replacing the vertex with the polygon formed by the intersections.
The number of nodes in a cubic graph is even, since the number of edges (vertices
* 3/2) must be an integer. For a Tait-coloring, just find the hamiltonian
cycle, alternate the colors in the cycle, and draw all remaining edges with
the third color. From this edge-coloring, the 4-coloring follows immediately,
as shown above. A beautiful idea, but it doesn't work!

Bill
Tutte was one of the mathematicians that cracked nazi codes. In 1946,
he started thinking that Tait's conjecture might be wrong (here's
a link to Tutte's exciting lecture). Eventually, he found the counterexample
that is 4-colored above. It is nonhamiltonian, and is thus a counterexample
to Tait's conjecture. One counterexample is enough to prove that a conjecture
is false.

Tait colorings are still interesting, though. For awhile, only
one cubic graph was known that could not be Tait-colored: the famous Petersen
graph. Graphs which could not be Tait-colored were called Snarks by Martin
Gardner, since like the Lewis Carroll beastie, they were hard to find. Snark
pictures.

The Petersen Graph (all of these graphs are isomorphic)

In 1966, Tutte noticed that all snarks were related to the Petersen
graph, and made the remarkable Tutte conjecture: Any Petersen-free bridgeless
cubic graph has a Tait-coloring. Petersen-free means that a Petersen
graph cannot be reached by deleting and/or contracting edges (a minor). Note
that the word "planar" is not mentioned. The 4 color theorem follows
as a consequence, since all planar graphs are Petersen-free.

In 2001, Robertson, Sanders, Seymour and Thomas found a proof
of Tutte's conjecture! The same team found the
improved 1996 proof of the four color theorem, but that can now be consider
a mere warm-up exercise - the Tutte conjecture required over 2000 different
unavoidable graphs, and more discharging rules than 4CT.

The below graph is the Coxeter
graph. It is non-planar, non-Hamiltonian, and symmetric, according to
the Foster Census.
For my math puzzle of the week, can this graph be Tait-colored? Or does it
have a Petersen minor? Is this a Snark? Send
Answer.

Erich Friedman:
I find it interesting that 1! 10! 22! 1! = 11! 0! 2! 21! . The same digits
broken up two different ways into factorials. I think this is the smallest
such example, but there must be many others. I've been looking for a number
that is the product of the factorials of its digits. Can anybody find one?
Send Answer.

Robert Heike spent 4 years developing a secret code, and used
it in an attempt to break out of prison. Deputy Debra Wesley of St. Lucie's
Sheriff department spent 90 minutes cracking it. Here
is Heike's letter. Here
is the press story. It's a very easy substitution cipher. Can you solve
it in less time? Answer.

Eric Weisstein (another astronomer) pointed me to the Bowl
of Integers, as something new at Mathworld. He's gotten over a thousand
emails since the return, and is very appreciative to all.

A well-known puzzle involves the game of Set. What is the largest
set of cards with no Set? The answer is at the
Set site. A variant of Set by Wei-Hwa Huang is Superset.
What is the largest set of cards with no Superset? Send
Answer. While playing around with a deck of cards numbered 1 to 60, I
started looking for subsets of three with a constant difference, like 1 7
13 (difference 6) or 4 27 50 (difference 23). What is the largest set of these
cards (1 to 60) that has no such subset? Send
Answer.

I'm still working on a puzzle involving special graphs of low
diameter. Maybe later tonight. Hey, now I can point you at Eric's listing
for the Coxeter Graph!
And many other wonderful things.

The above object is a rather special graph. Pick any two node
numbers. I claim that I'll be able to connect them with at most four other
nodes. For example 1 and 10. {1, 43, 44, 8, 9, 10}. The distance between 1
and 10 is 5. If you look at all the shortest paths between every set of nodes,
the longest such path is the Diameter of the graph. The above graph has an
amazingly low diameter.

I found the above graph at The
Foster Census. Sixty is a nice number, so I'm pondering what sorts of
card games could be built with it. But then, I found The
(Degree,Diameter) Problem for Graphs -- a page devoted to low diameter
graphs. I'm working on getting pictures of the big graphs, using Mathematica.
The graph below is the largest one know with a diameter of 4

I've been trying to get a picture for the 70 node graph, but
haven't figured it out, yet. I've also spent a few hours trying to find a
nicer representation of F60, above. It should have a more symmetrical representation,
but I haven't found it. I haven't perfected an F60 card game either.

Ted Drange is a Dan-level go player, and a master of small go
games. I met him through NOST (kNights
Of the Square Table). "Bill Spight pointed out to me that the alleged
canonical sequence that I give for 6x6 go (in the "Mini-go"
essay on the Mathpuzzle.com page) is defective. By proper play, White could
cut Black's win to only 1 point, instead of 4 points, starting at move #20,
but Black could thwart that by playing differently at move #15. An improved
(and I think the correct canonical) sequence is as follows:

The equilateral triangle is widely found in historic buildings
and structures across Europe.

In England there is a famous and unique Triangular Lodge at
Rushton inNorthamptonshire. It was built in 1593 by Sir Thomas Tresham as
anexpression of his Christian faith, and everywhere the number 3 is used tosymbolize
the Trinity (three-in-one: God the Father, Son and Holy Spirit).The groundplan
of the lodge is a perfect equilateral triangle, with eachside 33 feet long
(by tradition the age of Christ at his death). It hasthree floors, each floor
has three windows, and every window is a three-foldTrefoil. There are three
gables on each side, and 9 gargoyles. Even thecentral chimney is 3-sided.
It must be more than fifty feet tall (maybe 66feet!), and so this is a very
substantial stone and brick building. I visited it a few weeks ago and it
was very impressive.

There is a triangular castle at Gripsholm in Sweden; part of
the Chateau deChantilly in France is entirely based on an equilateral plan,
on a gigantic scale. Longford Castle near Odstock was built as a house in
the late 16th century and has an unusual triangular floor plan.

During the 18th century, several Folly Towers with triangular
foundations were built in England. The triangle provided a more cost-effective
way of achieving maximum height for what were, after all, essentially use-less
structures. These included Fort Belvedere in Windsor Great Park, the triangular
sham church on the Cotehele estate in Cornwall, and the Horton Tower known
as Sturts Folly at Horton, Dorset.

The former hedge maze at Arley Hall in Cheshire, whilst hexagonal
in outline, had an internal path geometry based on equilateral triangles;
I discovered this when seeking the most effective way of portraying its design
using a computer.

Approached up the only road with a hairpin bend in the Netherlands
(everywhere else is too flat!), Three Lands Point is the highest point in
the country, and the place where the borders of Germany, Belgium and the Netherlands
meet. Here in 1991, I created a giant hornbeam hedge maze 300 feet across,
with three distinctive pointed corners to its design, a central triangular
goal, three bridges, and 9 Foaming Fountain Gates. There is a colour maze
in the courtyard based on equilateral triangles, using the flags of these
three nations, each of which is portrayed by a flag containing three stripes
of different colours. Paths of three different colours are used to solve this
colour maze.

material added 30 October 2001

Is your name in the first 4 billion digits of Pi? At http://pi.nersc.gov/,
you can find out. I found "riddles" in Pi, but not "puzzle".

When I was 12 years old, I saw supermagnets mentioned in the
Edmund Scientific catalog. I saved up some money and bought one - a samarium
cobalt -- the strongest one I could afford. It arrived shortly before my family
left for a store, so I opened the box in the back of our 1970 Country Squire
station wagon. I experimentally started touching it to a steel plate in the
back of the car. Wham! My fingers were suddenly bashed and bleeding,
and the magnet was stuck there. My dad asked if anything was wrong. "Nothing,"
I think I said, in a lot of pain and shame. Still, I love magnetism, and I
did exploration with magnets this weekend. I levitated a piece of pyrolytic
graphite after seeing the experiment at www.scitoys.com.
Levitation is a lot of fun. I learned more as I searched around. Here is a
page describing levitated
frogs, and levitated sumo wrestlers. I picked up the graphite at Scitoys.
For the magnets, I finally opted for http://www.wondermagnets.com/.
They have pictures, good prices, good links, and good warning messages. No matter
what age you are, get one of the smaller magnets first, so that you can appreciate
their power. The Magnet Man site has a lot of good material. Another product I like, for $5, is
the set
of minimagnets.

The triangulated building has been tracked down. See www.federationsquare.com.au.
Thanks to David Eppstein
and Khalad Karim for tracking it down, and to Dick Hess for the original pictures.
If you know of any mathematical architecture, send it to me or David.

Dick Hess related the following trick: A
mathemagician has an audience select five numbers from the set {1,2,3...n}.
He thinks carefully about the numbers and then names four of the numbers and
asks an audience member to write those four numbers in that order on a blackboard.
(The audience contains no shills who might collude with the mathemagician).
The mathemagician then leaves the room and a partner who has not seen or heard
the proceedings so far is summoned to enter the room (without any contact
with the mathemagician). She studies the four numbers on the board and correctly
announces the fifth number from the original collection.

Comment: This has been discussed in several mathematical forums in recent
years; there are several known solutions for n=52 (I know of two, one being
handed out at G4G4 by Colm Mulcahy under the heading "Fitch Cheney's
Five Card Trick"). Considerable ingenuity is required to come up with
a practical scheme for the maximum value of n and the only known solution
requires some practice if your goal is to perform the trick. (I neither know
the maximum value for n for which the trick can be performed nor do I know
of any method for n>52). 52 Card Trick Solutions

Bill Cutler
has written a program that can solve my Kites &
Bricks puzzle (I still have a few available, at $15 each). It turns out
it has exactly 5 solutions. Koshi Arai and his
solving group found all five by hand about a year ago. Bill checked the desirable
7-7-7 case for me, and it turns out there are no solutions for it.

I was watching the TV show Alias on Sunday night (I like it),
and was surprised when it was revealed that Kate Jones was some sort of mastermind
international spy. She is certainly a puzzle mastermind, I'll vouch for that.
Kate runs Gamepuzzles.com, which
is the best toy store around for unusual puzzle gifts. Here is her article
on naming the 166 hexomino
pieces. Another good place to shop is Cleverwood.com
-- Kathleen Malcolmson can't claim to be an international spy, but she did
make some incredible deals while she was in Japan. She can now offer high
quality puzzle boxes at even lower prices.

The 40 cubes puzzle had a mistake! It was solved by Denis Borris,
Clinton Weaver and Jeffrey Smith. The typo was the k and K in lines 6 and
7. I give the corrected version below. (apologies for that).

FORTY CUBESby Rhombus

Koshi Arai, Mark Michell, Claudio
Baiocchi, Clinton Weaver, Liu
Wei, Juha Saukkola, and Joseph DeVincentis all solved my 17x17 domino-tromino
problem. Guenter Stertenbrink's program found sequential domino-tromino dissections
of the following rectangles: 2*5, 2*7, 7*10, 9*15, 10*12, 12*16, 14*14, 14*21,
14*31, 17*17, 24*29. He also created these problems: "Divide a 20x20
square into 11 different dominoes and trominoes. Divide a 17x17 square into
9 different dominoes and trominoes." Send
answers. These are my favorite types of problems - something short that
I can remember and solve when I have time.

material added 14 October 2001

From Erich Friedman -- cut a 20x20 square into 14 smaller squares
so that the biggest subsquare cannot be placed in the corner.

I started looking for possible variants on this. Let a Domino
be a n×2n rectangle, and a Tromino be a n×3n rectangle. If you
represent the series 1 to 6 with dominoes and trominoes, can a 14×14
square be made with the 6 pieces? Yes, it can. The solution is below. If you
represent the series 1 to 7 with dominoes and trominoes, can a 17×17
square be made with the 7 pieces? Yes, it can. A nice little puzzle.What is
the next square that can be covered with a sequential series of dominoes and
trominoes? I don't know, yet. Send Answers.

Lloyd King sent me a nice little puzzle he calls Die-Hard. Arrange
the six equilateral triangles to create a view of a regular die. If you can
figure out the numbers on the die, write
me. For another puzzle with dice, see Andy Brown's Puzzlemaniac
site (he also does dominoes and signal towers).

Die-Hard puzzle by Lloyd King

Eric Weisstein and I recently had the pleasure of hearing Clark
Kimberling give a lecture on triangles. You can get a taste of this at the
Encyclopedia of
Triangle Centers website (ETC). I've now been studying the mathematics
of trilinears
for several days now.

Part of the Slocum Puzzle Foundation Collection will be shown
in the Lilly Library at Indiana University Bloomington from October 11 to
December 18, 2001. Here are more details. Here's
a map (from MapQuest.com). I live in Champaign-Urbana, in the upper left.
I'll try to visit there October 20th. In related puzzle news, John Rausch
has updated Puzzle
World.

Erich Friedman's Math
Magic this month is about entangled polyforms. Chris Lusby Taylor has
improved the Picture Hanging solution. Koshi
sent me a picture from my visit to Tokyo. I'm in the Mathematica T-shirt.
A lot of people have wondered what I look like.

John Gowland wonders if the puzzles of Rhombus have been collected
anywhere. Rhombus is his favorite cross number constructor. He sent me the
following puzzle by Rhombus, as a sample. Forty Cubes is his all-time favorite.

I've become a big fan of Ken Krugler's Turmite
program. As soon as I find a few extra minutes of spare time, I'll show
Turmites discovered by Ken and myself. In the meantime, here is my current
Palm database file.

material added 30 September 2001

Can anybody pack all 12 pentominoes to a 9x9 square grid, so
that any pair of pentominoes share at most one grid segment? That is, two
adjacent pentominoes can only touch at one side of one square only. Jukka-Pekka
Ikäheimonen found a solution the day after he asked me the question.
The image shows only 11 pieces. Can you find JP's answer? This problem was
solved by Andy Brown, Matthew Prins, Jukka-Pekka Ikäheimonen,
Clint Weaver, Koshi Arai,
Juha Saukkola, Livio Zucca.

Samantha Levin, Denis Borris ("Boy,
Ed: what a beautiful challenging puzzle."), Jeffrey D. Smith ("The
puzzle was really good, so I hunted and found another that you had posted
on your site."), Bob Kraus ("a nice puzzle of its type"),
and Joseph DeVincentis ("I liked this puzzle.")
solved John Gowland's Four of a Kind problem (here's
the answer). Several people have asked me if I know of other places with
cross-number problems. One of them is at Arik Schenkler's site, at internetdollar.com.
Neil Fitzgerald solved the Hanging a Picture problem. Torsten Sillke kindly
sent me some history of this nice problem. (Write-ups)

material added 24 September 2001

It is possible to hang a painting with two nails so that removing
either of them will cause the picture to fall. Here's
how to do it. John Lawton found that this can be done with any number
of nails (remove one, and the painting falls). Can you see how to do it with
five or six nails? Answer.

I recently picked up a cheap Palm Pilot (a Handspring, actually).
These have gotten more powerful than I realized. One nice application for
it is a Wordplay
program by NPL member Kiran Kedlaya. I found a very nice "turmite"
program for the Palm OS ( Turmite1.0b
). A Windows version is at Kasprzyk's
ALife Page. A Linux version is call Ants and is included with the Gnome
packages.

Stan Wagon, in the latest MacPOW, notes there are exactly 13
unlucky numbers 2, 3, 5, 6, 7, 8, 12, 13, 14, 15, 19, 21, and 23. A lucky
number is one that can be partitioned so that the sum of its reciprocals is
exactly 1. For example, 11 = 2+3+6 is a partition of 11 with the reciprocals
of the terms summing to 1. He created algorithm that can find all the lucky
partitions of 60 in less than 4 seconds, and challenges others to do likewise.
(Out of curiousity, how many of my readers have picked up Stan Wagon's The
Mathematical Explorer? Please write
me with your thoughts about it, and I'll send you something interesting.
Don't miss it at the introductory price of $70.)

John Gowland sent me a new logic puzzle. In this crossnumber,
p and q are three digit primes whose product is a 5 digit number,
which can be considered as a "four of a kind" in a poker hand. e.g.
113 x 823 = 92999. This product is clued as a 3-digit number r according
to the following convention: A number in the form 12222 will be clued as 122;
a number in the form 11112 will be clued as 112; and numbers in the form 12111,
11211, or 11121 will all be clued as 121. Across entries are denoted by capitals
and down entries are denoted by small letters. There are no zeroes in the
completed diagram. Answer.

Brendan Owen and Patrick Hamlyn put together the following checkered
octiamond solution:

material added 15 September 2001

I met Brendan
Owen via Christopher Monckton and the Eternity
Puzzle. This was a spectacular risk on Christopher's part, to make a fair
puzzle that gave a million pounds to the first solver. It led to new refinement
of the Breadth First Search algorithm, which will be of major importance eventually.
Brendan was part of the research team known as the Eternity Mailing List.

Brendan, an Australian, hurt his foot
while rock climbing four years ago. Early this year, while visiting the UK,
he got a very bad infection in his foot. His father looked through Brendan's
email and noticed that Martin
Watson lived near the hospital, and contacted him. I spoke with Martin
while I was in Tokyo. For several weeks, Martin visited Brendan at the hospital.
Now, Brendan has mostly recovered, but needs to keep his leg elevated 95%
of the time, which means a lot of time at the computer.

Adrian Fisher collects books on interesting
architecture, and here's one that sounds fascinating: "Castel del Monte,
Geometric Marvel of the Middle Ages" by Heinz Gotze, published by Prestel,
Munich and New York, in 1998; ISBN 3 7913 1930 2. This entire 200 page large
format illustrated book is seemingly devoted to a single octagonal castle
in Southern Italy. In fact, the astonishing octagonal geometry of this castle
is merely the catalyst for a complete celebration of the Octagon and its related
geometry, from octagonal Roman mosaics and tiling patterns in the Alhambra
Palace in Spain, to Jerusalem's Dome of the Rock, and the Church of San Vitale
in Ravenna (which also contains a splendid multi-coloured marble pavement
labyrinth). The accuracy of stone craftsmanship of the castle in three dimensions
is awesome. The geometry of the Castle itself is astonishing - a central octagonal
core containing an inner octagonal courtyard, surrounded by 8 perfectly octagonal
towers, all rising to a great height more in the character of a cathedral
(or a triumphal palace) than a defensive place of refuge. Certainly the castle
remains undamaged to this day, as if it never experienced hostilities."
Now that I know the name, I found a lot of info about the castle on the internet.

material added 12 September 2001

Stan
Wagon has posted problem of the week 939 to the macpow mailing list :
Find 939 distinct positive integers such that for every seven of them, their
product is divisible by their sum. The answer to Andy Liu's Lion and Lamb
problem was also posted.

Juha Saukkola, Joseph DeVincentis, and Don Woods found the distribution
of Urns that will cover all possible distributions of draws: 001122, 000012,
111102, 222201. Robert Jenkins has a score of 18 on the Don Woods challenge.

On Monday, I received a Polymorf
set. It's a very good set, and reasonably priced. I like Zome,
too, but Polymorf allows an easy study of polyhedral nets. Fascinating.

material added 11 September 2001
-- Addendum

Nothing can stop the wonder of mathematics. It has survived
time, war, attacks, and disasters. St.
Andrew lists the following about Archimedes.

Archimedes ... was ..., as fate would have it, intent upon working
out some problem by a diagram, and having fixed his mind alike and his
eyes upon the subject of his speculation, he never noticed the incursion
of the Romans, nor that the city was taken. In this transport of study
and contemplation, a soldier, unexpectedly coming up to him, commanded
him to follow to Marcellus; which he declining to do before he had worked
out his problem to a demonstration, the soldier, enraged, drew his sword
and ran him through.

The problem was more important than the war, to Archimedes.
Today, he and his work are still revered. St.
Andrews has many such stories of mathematicians. Sophus
Lie made some of his greatest discoveries while in prison, suspected of
being a spy. During my last visit with Stephen Wolfram, we pulled out an older
book of Integral tables as a reference, and it turned out to have been the
possession of a WWII German prisoner-of-war in a US camp. Adrian Fisher, a
maze maker and my friend, lives on a street that was heavily bombed during
that war. Nob Yoshigahara, a survivor of both a large chemical explosion and
cancer, is one of the great puzzle creators and ambassabors. He personally
welcomed me to Japan. Vladeta Jovovic, a programmer from Belgrade, Yugoslavia,
sent me a nice recreational math program while
his country was getting bombed by mine.

I spent some of the day working on a diagram of Stephen's based
on Euclid's Elements. This was a favorite book of Abraham Lincoln, who memorized
many of the proofs. One of my co-workers pointed out that the image I posted
earlier today was very similar to Voronoi
diagrams. I met Eric Weisstein's father, and showed him a puzzle based
on one of Eric's diagrams.

material added 2 September 2001

The entries for the first International
Puzzle Party Design Competition have been put together by Nick Baxter.
There are many excellent puzzles here, of course. Of particular mathematical
interest are ... oh ... half of them. Many of the puzzles are quite affordable.
As a puzzle for here, Nick would like you to look at the Yosegi
puzzle. It consists of 32 half-square right triangles. 16 of them are
light-dark-light, and 16 are dark-light-dark. Nick's puzzle: How many designs
are possible? Slightly harder: How many designs are possible without similar
colors touching? Send answer.

Roger Phillips has sent me a way to tile a 7x7 torus with 7
heptominos so that they all touch each other. In fact, he sent me 18 different
solutions. He also found 22 different hexominoes that will cover a 6x6 square,
so that all touch each other. Here's an image suitable
for wallpaper. Mark Thompson, Guenter Stertenbrink, and Roel Huisman also
worked on this problem.

Dan Blum and Jim Boyce found that 4 urns is enough for Don Woods
problem below. Can you find the distribution? Send
Answer. Robert Jenkins has managed to get a score of 18 on the Don Woods
challenge.

An immense ammount of good stuff is available at www.g4g4.com,
including a full book.

Marcel Tuenissen and Ramon van der Winkel have examined Dr.
Wood's iamond problem (way at the bottom). The found there are exactly
2035428 ways to fill this shape with the checkered hexiamonds. My own puzzle:
If the Octiamonds are checkered in all possible ways, you wind up with 225
pieces. Can they make a side-30 checkered diamond?

A pretty proof of mine. I know Alexander Soifer published this
in the USSR many years ago, whilst still a lad. The problem: the sides of
a triangle are trisected, and lines are drawn from the corners to three of
the trisection marks. What is the area of the internal triangle that is created,
compared with the original triangle?

material added 26 August 2001

Don Woods, one of
the creators of the original Adventure game, has kindly sent me his new Twenty
Questions puzzle. Send Answer.
He constructed it after seeing Jim Propp's Self-Refential
Aptitude test. Don also described a different puzzle he is working on.
Let T be the the number of object types (if T=3, there are 3 types of object),
and let D be the number of objects that can be drawn (If D=4, you will draw
4 objects). In the case T=3, D=4, the possible draws are: 0000, 0001, 0002,
0011, 0012, 0022, 0111, 0112, 0122, 0222, 1111, 1112, 1122, 1222, and 2222.
Let U be number of objects an Urn can hold. Suppose that U=6. You need at
least 5 urns to cover all possible draws: 011112, 112222, 001222, 000112,
and 000022 will do. Can anyone find a method for finding the number of Urns
required to cover all possible draws, given T, D and U? This is related to
items in the La Jolla Covering
Repository, but is a different problem.

A photo by Dick Hess, taken in Melbourne (Flinders street). Same idea as Kites
& Bricks.

The place where I work, Wolfram Research,
has launched The
Mathematical Explorer (TME). It's a $69 program that contains
an interactive book in recreational mathematics by Stan Wagon (a long-time
#1 writer on my books page). Mathematica is
not needed to run TME. Instead, the TME has a self-contained version of Mathematica
with certain high level functions disabled, and new recreational functions
added. It does a lot.

material added 19 August 2001

Before landing in Tokyo, I had to fill
out a disembarkation form. This was the first time I was able to list my occupation
as "Mathematician." I figured out the train system quickly enough.
I tried a triangle of sticky rice wrapped in a seaweed paper, and found it
was very good. It had tuna at the center for flavoring. Nob Yoshigahara helped
me identify other flavors. They were all good, and cost 100 yen (about 80
cents). Nob is letting me share an idea of his. One day, he pondered the dodecahedron.
12 faces. There are 24 ways that 1,2,3,4,5 can be arranged in a circle. Each
such circle is a mirror image of one of the others. He put these mirror images
on either side of a pentagon, to obtain 12 pentagons. (1+2+3+4+5)*12 = 180
= 9*20. Can these pentagons be arranged so that all 20 corners add up to 9?
Nob solved it. The below figure has all 12 permutations of {1 2 3 4
5}, and each corner has a sum of 9. It's neat. Things like this make life
more fun.

Oskar van Deventer tried out a more elaborate
japanese meal with me and some others. Oscar had several new puzzle creations
with him. One of the favorites was his velcro blocks
puzzle, that he detailed earlier. This is a great little maze, and quite
challenging. It's the next thing I'll make for myself. I bought a copy of
his Metro puzzle,
and like it a lot. Another puzzle he gave me permission to mention involved
a velcro covered icosahedron, and a velcro strip of 23 triangles. The task
was to completely cover the icosahedron with the strip. Very nice puzzle,
and easy to make. He's examined all the deltahedra and zonohedra as well

Alan Segal, manager of Bits
and Pieces, gave me a lovely anodized aluminum burr puzzle. He had a full
collection of metal puzzles available there, and I bought one of each. He
personally asked me to invite everyone to see the range of metal puzzles he
has available. Many of the brass puzzles were designed by Rocky
Chiaro. Rocky sold me one of his hand-made brass creations. Others are
available at his website. Also looking at metal, I met the owner of a
site devoted to ancient
chinese locks.